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iop360

  • 3 years ago

Find all distinct roots(real or complex) of z^2 + (9+3i)z + (24+11i) in the form a+bi. Hint: You may have to complete the square

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  1. iop360
    • 3 years ago
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    http://answers.yahoo.com/question/index?qid=20121124223053AAchfPc the bottom explanation of this link might be it, but i dont understand it

  2. iop360
    • 3 years ago
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    have you figured anything out?

  3. sirm3d
    • 3 years ago
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    \[\large z^2+(9+3i)z+(24-11i)=0\]\[\large z^2+(9+3i)z=-(24-11i)\]to complete squares, add (b/2)^2 to both sides, where b=9+3i \[\large z^2+(9+3i)z+\left( \frac{ 9+3i }{ 2 } \right)^2=-(24-11i)+\left( \frac{ 9+3i }{ 2 } \right)^2\]

  4. iop360
    • 3 years ago
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    yeah im at this step... what do you do next?

  5. iop360
    • 3 years ago
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    \[(z + (\frac{ 9 + 3i }{ 2 }))^2 = -24 + 11i + 18 +\frac{ 27i }{ 2 }\]

  6. iop360
    • 3 years ago
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    \[z + (\frac{ 9+3i }{ 2 }) = \pm \sqrt{\frac{ -12 + 5i}{ 2 }} \]

  7. iop360
    • 3 years ago
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    i dunno, lol

  8. iop360
    • 3 years ago
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    i dont get how the person completed the square in the yahoo link..

  9. iop360
    • 3 years ago
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    \[z = -(\frac{ 9+3i }{ 2}) \pm \sqrt{\frac{ -12 + 5i }{ 2 }}\]

  10. iop360
    • 3 years ago
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    it would be so much easier if the square root wasnt there...we could just add them up easily

  11. iop360
    • 3 years ago
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    from the yahoo answers link: z² + (2−4i)z = (11+10i) z² + (2−4i)z + (-3 -4i) = (11+10i) + (-3 -4i)

  12. iop360
    • 3 years ago
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    what i dont get is where the (-3-4i) comes from

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